7.3 Barrier Sweep Coverage for a finite Curve
7.3.2 Energy restricted barrier sweep coverage
In this section, we propose an algorithm for the energy restricted barrier sweep coverage problem. The approximation factor of the proposed algorithm is 133 though it is not known whether the problem is NP-hard or not. Lete be the energy source in the plane.
To make the problem feasible, we assume that the distance of any point on C from e is less than vΓ2 .
Definition 7.3.1. (e-tour) A tour, denoted by {e, p, q, e}is called an e-tour if it starts from e, visits arc(pq) from p to q along C and then returns to e such that total length of the tour is at most vΓ, where p and q are two points on C.
For example {e, i1, i2, e} is an e-tour as shown in Fig. 7.1, provided the length of the tour is at most vΓ.
The objective of our technique is to find a tour through e and C, which is concate- nation of multiple number of e-tours. Letd(a, b) be the Euclidean distance between two points a and b. Let dc(p, q) be the distance between two points p and q along C in the clockwise direction, wherepandq are two points onC. So,dc(p, q) is equal to the length of the arc(pq).
i1
i2
i3
i4
e
Figure 7.1: Showing selection of e-tours {e, i1, i2, e},{e, i2, i3, e}, {e, i3, i4, e},· · · First we consider C is a finite ‘closed’ curve and solve Problem 10 for it. Later we describe how to solve the problem for an ‘open’ curve. We choose any point i1 on
C. Find a point i2 on C in the clockwise direction from i1 (Ref. Fig. 7.1) such that Algorithm 7 EnergyRestrictedBSC
1: Choose any point i1 onC.
2: C0 =C,n0 = 1.
3: while C0 6=φ do
4: if d(e, in0) +|C0| ≤ vΓ2 then
5: h=i1.
6: else
7: Select a point h on C in a clockwise direction from in0 such that d(in0, h) =
vΓ
2 −d(e, in0).
8: end if
9: if n0 6= 1 andd(e, in0−1) +dc(in0−1, h) +d(e, h)≤vΓthen
10: C0 =C0\arc(in0h).
11: in0 =h, Tn0−1 ={e, in0−1, in0, e}.
12: else
13: in0+1 =h,Tn0 = {e, in0, in0+1, e}.
14: C0 =C0\arc(in0in0+1).
15: n0 =n0+ 1.
16: end if
17: C0 =C0 \arc(in0in0+1).
18: end while
19: AP P RX =T1· T2· T2· · · Tn0, where ‘·’ is denoted as concatenation operation.
20: Divide AP P RX into equal parts of length vt and deploy one mobile sensor at each of the partitioning points.
21: All mobile sensor start moving along AP P RX at same time in same direction.
dc(i1, i2) = vΓ2 −d(e, i1). Here T1 ={e, i1, i2, e} is an e-tour, since length of T1 is equal to d(e, i1) +dc(i1, i2) +d(i2, e)≤d(e, i1) + vΓ2 −d(e, i1) +vΓ2 = vΓ, as shown in the Fig.
7.1. Next e-tour is selected as {e, i2, i3, e}, where i3 is a point on C in the clockwise direction from i2 such that dc(i2, i3) = vΓ2 −d(e, i2). Once the second tour is selected, we check whether the combination of the previous tour and current tour together, i.e., {e, i1, i3, e} forms an e-tour or not. If the combined tour {e, i1, i3, e} is a valid e-tour, then the previous tour {e, i1, i2, e} is updated into {e, i1, i3, e} and we proceed to select the nexte-tour. This updating process continues until combined tour violates the length constraint ofe-tour. In general, after computing an e-tourTj ={e, ij, ij+1, e}, we select next point ij+2 onC such thatdc(ij+1, ij+2) = vΓ2 −d(e, ij+1). Then if {e, ij, ij+2, e}this is a valid e-tour, we update the tour Tj as Tj ={e, ij, ij+2, e}. Otherwise the tour Tj+1
is selected as Tj+1 ={e, ij+1, ij+2, e}. Finally C is decomposed into multiple number of e-tours such that every point on C is included in some e-tour.
If C is an ‘open’ curve, we use the above technique to decompose C into multiple number of e-tours considering one end point of C as i1 and continue till the other end point. Note that according to the construction of the e-tours, length of the combined tour of any two consecutive e-tours is always greater than vΓ.
Let AP P RX be the total tour after concatenation of the e-tours, T1,T2,T3,· · · one after another in order of obtained by the above technique. For exampleT1 ={e, i1, i2, e}, T2 = {e, i2, i3, e}, T3 = {e, i3, i4, e},· · · are e-tours as shown in the Fig. 7.1, where concatenation of those tours is T1· T2· T3· · · · ={e, i1, i2, e} · {e, i2, i3, e} · {e, i3, i4, e} · · · ·
={e, i1, i2, e, i2, i3, e, i3, i4, e,· · · }, where ‘·’ is denoted as the concatenation operation of the e-tours. Let|AP P RX|be the length ofAP P RX. Divide AP P RX into equal parts of lengthvtand deploy one mobile sensor at each of the partitioning points. The mobile sensors then start moving alongAP P RX in the same direction to ensure sweep coverage of C. The Algorithm 7 (EnergyRestrictedBSC) for energy restricted barrier sweep coverage is proposed for a closed curve. But as explained above it is also applicable for an open curve.
Theorem 7.3.2. According to Algorithm 7, each mobile sensor visits e in every Γ time period and each point onC is visited by at least one mobile sensor in every ttime period.
Proof. Mobile sensors move along the tourAP P RX according to movement strategy of Algorithm 7. As length of each e-tour is less than or equals to vΓ, the mobile sensors visiteafter traveling at mostvΓ distance since its last visit ofe. Therefore, each mobile sensor visits e once in every Γ time period. Again, the mobile sensors are deployed at every partitioning points ofAP P RX and two consecutive partitioning points are within vt distance apart. The relative distance between any two consecutive mobile sensors is at most vt at any time as they are moving in same speed v and same direction. So, when any point p onC is visited by a mobile sensor at t0, another mobile sensor was on the way topand within the distance ofvtalongAP P RX. Hence pwill be again visited by another mobile sensor within next t time.
To analyze the approximation factor of Algorithm 7, we consider some special points
on C. Let i1p, i2p be two special points on the arc(ipip+1) of C such that the arc(ipip+1) is partitioned into three equal parts, i.e., the length of arc(ipi1p) equal to the length of arc(i1pi2p) equal to the length of arc(i2pip+1). We define a set of points, I ={ij, i1j, i2j|j = 1 to n0}. Following two lemmas give an upper bound of the length of the tour AP P RX and a lower bound of the length of the optimal tour respectively.
Lemma 7.3.3. |AP P RX| ≤ 1
3 2X
j
d(e, ij) +d(e, i1j) +d(e, i2j)
+ 5|C|
! . Proof. According to Algorithm 7, total length of the tour AP P RX is
|AP P RX| = d(e, i1) +dc(i1, i2) +d(i2, e) +d(e, i2) +dc(i2, i3) +d(i3, e) +· · ·+d(e, i1)
= |C|+ 2X
j
d(e, ij). (7.3.1)
Now by triangle inequality,
d(e, ij) ≤ d(e, i1j) +dc(ij, i1j) and d(e, ij) ≤ d(e, i2j) +dc(ij, i2j).
Therefore,
|AP P RX| ≤ |C|+ 2X
j
d(e, i1j) +dc(ij, i1j)
. (7.3.2)
|AP P RX| ≤ |C|+ 2X
j
d(e, i2j) +dc(ij, i2j)
. (7.3.3)
From the above Equation 7.3.1, 7.3.2 and 7.3.3, we can write, 3|AP P RX| ≤ 3|C|+ 2X
j
d(e, ij) + 2X
j
d(e, i1j) +dc(ij, i1j) +2X
j
d(e, i2j) +dc(ij, i2j)
. (7.3.4)
As the points i1j and i2j divide length of thearc(ijij+1) into three equal parts, therefore, X
j
dc(ij, i1j) = |C|
3 and X
j
dc(ij, i2j) = 2|C|
3 .
Using these two results, Equation 7.3.4 can be written as:
3|AP P RX| ≤ 3|C|+ 2X
j
d(e, ij) +d(e, i1j) +d(e, i2j)
+ 2|C|
3 +4|C|
3
|AP P RX| ≤ 1
3 2X
j
d(e, ij) +d(e, i1j) +d(e, i2j) + 5|C|
! .
p
ij
ij+1
ij−1
q
ij+2
e
Figure 7.2: Showing onee-tour OP Tl ={e, p, q, e}of the optimal tour OP T
Lemma 7.3.4. LetOP T be the optimal tour of the mobile sensors and |OP T| be length of OP T, then |OP T| ≥max
(1 4
X
j
d(e, ij) +d(e, i1j) +d(e, i2j) ,|C|
) .
Proof. Let OP Tl = {e, p, q, e} be an e-tour of the optimal tour OP T (Ref. Fig. 7.2).
We claim that at most one arc(ijij+1), part of an e-tour computed using Algorithm 7 for some j, can be completely contained in arc(pq), where arc(pq) is a part of OP Tl. To prove this, let us assume that there are two such arcs arc(ij−1ij) and arc(ijij+1) completely contained inarc(p, q). As length of combined tour of two consecutive e-tours is always greater than vΓ according to step 12 of Algorithm 7, therefore,
(d(e, ij−1) +dc(ij−1, ij+1) +d(e, ij+1))> vΓ. (7.3.5)
Now,
|OP Tl| = d(e, p) +dc(p, ij−1) +dc(ij−1, ij) +dc(ij, ij+1) +dc(ij+1, q) +d(e, q)
≥ d(e, ij−1) +dc(ij−1, ij+1) +d(e, ij+1)
> vΓ (from Equation 7.3.5).
This contradicts the fact that OP Tl is ane-tour. Therefore, at most one arc(ijij+1) for somej, can be completely contained inarc(pq). Hence, at most one complete arc(ijij+1) and twoarc(ij−1ij) andarc(ij+1ij+2) partially contained inarc(pq) as shown in Fig. 7.2.
Therefore, maximum eight points from the set I may belong to thearc(pq), since there are four points ij, i1j, i2j, ij+1 for the arc(ijij+1) and at most four special points, i1j−1, i2j−1 and i1j+1, i2j+1 for the arc(ij−1ij) andarc(ij+1ij+2) respectively.
Now, for any point x onarc(pq) ofOP Tl implies|OP Tl| ≥2d(e, x). As there are at most eight points of I in arc(pq), which implies
|OP Tl| ≥ 2P
x ∈ I∩arc(pq)d(e, x)
8 . (7.3.6)
Since all the points in I are on arc(pq) for some OP Tl, where OP Tl is a part of OP T, therefore,
|OP T| ≥ X
l
|OP Tl| ≥ 2P
j d(e, ij) +d(e, i1j) +d(e, i2j)
8 . Also, |OP T| ≥ |C|. Hence, |OP T| ≥ max
(1 4
X
j
d(e, ij) +d(e, i1j) +d(e, i2j) ,|C|
) .
Theorem 7.3.5. The approximation factor of Algorithm 7 is 13 3 .
Proof. Let N be the number of mobile sensors needed in our solution and Nopt be the number of mobile sensors in the optimal solution.
Then N =
|AP P RX| vt
and Nopt ≥ |OP T|
vt . From Lemma 7.3.3 and Lemma 7.3.4, we have
N
Nopt ≤ |AP P RX|
|OP T| ≤ 8 3 + 5
3 = 13
3 . Hence the approximation factor of our proposed Algorithm 7 is 13
3 .